I'm wondering if there is any information on the inverse of the Riemann zeta function (not it's reciprocal, but its functional inverse). This would obviously be a multivalued function.
closed as too broad by Myshkin, Franz Lemmermeyer, Stefan Waldmann, Alex Degtyarev, Ben Linowitz Oct 23 '16 at 17:21
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1$\begingroup$ What kind of information? $\endgroup$ – Qiaochu Yuan Aug 9 '11 at 23:39

34$\begingroup$ Supposedly it takes some constrained values at $0$. $\endgroup$ – Moosbrugger Aug 9 '11 at 23:39

5$\begingroup$ There are several books on the Riemann zeta function. Tell us which ones you've looked through, so we don't duplicate the effort you've already put in. $\endgroup$ – Gerry Myerson Aug 9 '11 at 23:40

$\begingroup$ @Gerry: I haven't looked at any. I was just curious about this topic and I googled it, not finding anything even remotely related (when people say inverse, they usually mean reciprocal). I'm looking for papers or books that have considered constructing the branches of the inverse function. $\endgroup$ – Victor Liu Aug 10 '11 at 0:33
I fiddled with this recently; however I did not yet arrive at an interesting result for the inverse of the $\zeta$. But one could use the alternatingzeta (or Dirichlet's eta $\eta$) function. It is not too difficult to construct an invertible powerseries for the related etafunction; one can simply consider the sequence of formal powerseries for $ {1 \over (1+1)^x }, {1 \over (1+2)^x}, ... $ and adds the coefficients at the same powers of x. These produce nonconvergent series, but which can be Eulersummed. In Pari/GP one does simply:
ps_eta = sumalt(k=0,(1)^k*taylor(1/(1+k)^x,x))
and gets
$ \small pseta(x)= 0.500000 + 0.225791 x  0.0305146 x^2  0.00391245 x^3 + 0.00208483 x^4  0.000312274 x^5 + O(x^{6}) $
From this a powerseries for zeta is also constructible:
ps_zeta = ps_eta /(12*2^x)
printtex(ps_zeta+taylor(1/(1x),x))
$ \small \begin{array} {lll} pszeta(x) &=&  0.500000 + (0.08106151) x  (0.00317823+1) x^2  (0.000785194+1) x^3 \\\ & & + (0.0001207001) x^4  (0.00000194090+1) x^5 + O(x^{6}) \end{array} $
(That powerseries is related to the power series using the Stieltjesconstant by replacing x by x+1 )
The power series for $\eta$ can be recentered at the fixpoint $ \small fp \sim 0.629334 $ to get a powerseries without a constant term which can then be inverted. Let's call this $\eta_{fp} = \eta (x+fp)fp $ then the powerseries begins like
$ \small \begin{array} {lll} pseta_{fp}(x) &=& 0.184574 x  0.0337023 x^2 + 0.000152965 x^3 + 0.00117594 x^4 \\\ & &  0.000254950 x^5 + 0.0000216757 x^6 + 0.00000147274 x^7  0.000000714222 x^8 + O(x^{9}) \end{array} $
From this we can generate a powerseries for the inverse of $ \eta_{fp}$. The range of convergence is small; but using eulersummation one can compute values for the inverse of $ \eta_{fp}$. Even fractional iterates are accessible; here is a plot which shows the fractional iteration of $\eta(x,h)$, beginning at x=1 where h is the iterationparameter (all is computed using the centered version $\eta_{fp}$ ).
The plot has to be read that at h=0 we have $\eta(x,0)=x=1 $, at h=1 we have $\eta(x,1)=\eta(x)= \log(2)$, at h=2 we have $\eta(x,2)=\eta(\eta(x))$ at h>inf we get the fixpoint fp and the inverse is at h=1: $ \eta(x,1)=\eta^{1}(x) \to \infty $
I'm not yet ready with a small script/sketch of an article where I explore this in a bit more detail.
However, as I said, the inverse of zeta does not behave so nicely  the eigenvalue for the iteration is negative and one cannot uniquely get fractional roots out of this. Also the according powerseries may be too bad configured/has too small range of convergence.
[update] Here is a plot for a range of the inverse alternating zeta; for the "extreme" values at the borders I used Eulersummation because the power series has very small range of convergence.
The question is about the value distribution of $\zeta(s)$; it is considered (without speaking of inverse) in some detail in Chapter XI of Titchmarsh's book.

1$\begingroup$ Presumably amazon.com/TheoryRiemannZetaFunctionSciencePublications/dp/… $\endgroup$ – Todd Trimble♦ Aug 10 '11 at 16:42